1 STEP : Let's replace the function
w ( z ) = y ( z ) ⋅ z 1 / 4 → { w z ′ = y ′ ⋅ z 1 / 4 + 1 4 ⋅ y ⋅ z − 3 / 4 w z z ′ ′ = y ′ ′ ⋅ z 1 / 4 + 2 4 ⋅ y ′ ⋅ z − 3 / 4 − 3 16 ⋅ y ⋅ z − 7 / 4 → { w z ′ = y ′ ⋅ z 1 / 4 + y 4 z 3 / 4 w z z ′ ′ = y ′ ′ ⋅ z 1 / 4 + y ′ 2 z 3 / 4 − 3 y 16 z 7 / 4 → z 2 ⋅ w z z ′ ′ + 3 16 ( z + 1 ) w = 0 → z 2 ⋅ ( y ′ ′ ⋅ z 1 / 4 + y ′ 2 z 3 / 4 − 3 y 16 z 7 / 4 ) + 3 16 ⋅ ( z + 1 ) ⋅ y ⋅ z 1 / 4 = 0 z 9 / 4 ⋅ y ′ ′ + z 5 / 4 ⋅ y ′ 2 − 3 16 ⋅ z 1 / 4 ⋅ y − 3 ( z + 1 ) 16 ⋅ z 1 / 4 ⋅ y = 0 z 9 / 4 ⋅ y ′ ′ + z 5 / 4 ⋅ y ′ 2 − 3 ( z + 1 − 1 ) 16 ⋅ z 1 / 4 ⋅ y = 0 z 9 / 4 ⋅ y ′ ′ + z 5 / 4 ⋅ y ′ 2 − 3 16 ⋅ z 5 / 4 ⋅ y = 0 ∣ ÷ ( z 5 / 4 ) z ⋅ y ′ ′ + y ′ 2 − 3 16 ⋅ y = 0 w(z)=y(z)\cdot z^{1/4}\to\\[0.3cm]
\left\{\begin{array}{l}
w'_z=y'\cdot z^{1/4}+\displaystyle\frac{1}{4}\cdot y\cdot z^{-3/4}\\[0.3cm]
w''_{zz}=y''\cdot z^{1/4}+\displaystyle\frac{2}{4}\cdot y'\cdot z^{-3/4}-\displaystyle\frac{3}{16}\cdot y\cdot z^{-7/4}\\[0.3cm]
\end{array}\right.\to\\[0.3cm]
\left\{\begin{array}{l}
w'_z=y'\cdot z^{1/4}+\displaystyle\frac{y}{4z^{3/4}}\\[0.3cm]
w''_{zz}=y''\cdot z^{1/4}+\displaystyle\frac{y'}{2z^{3/4}}-\displaystyle\frac{3y}{16z^{7/4}}\\[0.3cm]
\end{array}\right.\to\\[0.3cm]
z^2\cdot w''_{zz}+\frac{3}{16}(z+1)w=0\to\\[0.3cm]
z^2\cdot\left(y''\cdot z^{1/4}+\frac{y'}{2z^{3/4}}-\frac{3y}{16z^{7/4}}\right)+\frac{3}{16}\cdot(z+1)\cdot y\cdot z^{1/4}=0\\[0.3cm]
z^{9/4}\cdot y''+z^{5/4}\cdot\frac{y'}{2}-\frac{3}{16}\cdot z^{1/4}\cdot y-\frac{3(z+1)}{16}\cdot z^{1/4}\cdot y=0\\[0.3cm]
z^{9/4}\cdot y''+z^{5/4}\cdot\frac{y'}{2}-\frac{3(z+1-1)}{16}\cdot z^{1/4}\cdot y=0\\[0.3cm]
\left.z^{9/4}\cdot y''+z^{5/4}\cdot\frac{y'}{2}-\frac{3}{16}\cdot z^{5/4}\cdot y=0\right|\div\left(z^{5/4}\right)\\[0.3cm]
\boxed{z\cdot y''+\frac{y'}{2}-\frac{3}{16}\cdot y=0}\\[0.3cm] w ( z ) = y ( z ) ⋅ z 1/4 → ⎩ ⎨ ⎧ w z ′ = y ′ ⋅ z 1/4 + 4 1 ⋅ y ⋅ z − 3/4 w zz ′′ = y ′′ ⋅ z 1/4 + 4 2 ⋅ y ′ ⋅ z − 3/4 − 16 3 ⋅ y ⋅ z − 7/4 → ⎩ ⎨ ⎧ w z ′ = y ′ ⋅ z 1/4 + 4 z 3/4 y w zz ′′ = y ′′ ⋅ z 1/4 + 2 z 3/4 y ′ − 16 z 7/4 3 y → z 2 ⋅ w zz ′′ + 16 3 ( z + 1 ) w = 0 → z 2 ⋅ ( y ′′ ⋅ z 1/4 + 2 z 3/4 y ′ − 16 z 7/4 3 y ) + 16 3 ⋅ ( z + 1 ) ⋅ y ⋅ z 1/4 = 0 z 9/4 ⋅ y ′′ + z 5/4 ⋅ 2 y ′ − 16 3 ⋅ z 1/4 ⋅ y − 16 3 ( z + 1 ) ⋅ z 1/4 ⋅ y = 0 z 9/4 ⋅ y ′′ + z 5/4 ⋅ 2 y ′ − 16 3 ( z + 1 − 1 ) ⋅ z 1/4 ⋅ y = 0 z 9/4 ⋅ y ′′ + z 5/4 ⋅ 2 y ′ − 16 3 ⋅ z 5/4 ⋅ y = 0 ∣ ∣ ÷ ( z 5/4 ) z ⋅ y ′′ + 2 y ′ − 16 3 ⋅ y = 0
2 STEP : Let's change the variable
z = x 2 → y x ′ = d y d x = d y d z ⋅ d z d x = y ′ ⋅ 2 x y ′ = y x ′ 2 x y x x ′ ′ = d d x ( y x ′ ) = d d x ( y ′ ⋅ 2 x ) = 2 x ⋅ d ( y ′ ) d z ⋅ d z d x + y ′ ⋅ 2 y x x ′ ′ = y ′ ′ ⋅ 4 x 2 + y x ′ 2 x ⋅ 2 → y ′ ′ = 1 4 x 2 ( y x x ′ ′ − y x ′ x ) z=x^2\to y'_x=\frac{dy}{dx}=\frac{dy}{dz}\cdot\frac{dz}{dx}=y'\cdot2x\\[0.3cm]
\boxed{y'=\frac{y'_x}{2x}}\\[0.3cm]
y''_{xx}=\frac{d}{dx}\left(y'_x\right)=\frac{d}{dx}\left(y'\cdot 2x\right)=2x\cdot\frac{d\left(y'\right)}{dz}\cdot\frac{dz}{dx}+y'\cdot2\\[0.3cm]
y''_{xx}=y''\cdot 4x^2+\frac{y'_x}{2x}\cdot2\to\boxed{y''=\frac{1}{4x^2}\left(y''_{xx}-\frac{y'_x}{x}\right)} z = x 2 → y x ′ = d x d y = d z d y ⋅ d x d z = y ′ ⋅ 2 x y ′ = 2 x y x ′ y xx ′′ = d x d ( y x ′ ) = d x d ( y ′ ⋅ 2 x ) = 2 x ⋅ d z d ( y ′ ) ⋅ d x d z + y ′ ⋅ 2 y xx ′′ = y ′′ ⋅ 4 x 2 + 2 x y x ′ ⋅ 2 → y ′′ = 4 x 2 1 ( y xx ′′ − x y x ′ )
z = x 2 → y x ′ = d y d x = d y d z ⋅ d z d x = y ′ ⋅ 2 x y ′ = y x ′ 2 x y x x ′ ′ = d d x ( y x ′ ) = d d x ( y ′ ⋅ 2 x ) = 2 x ⋅ d ( y ′ ) d z ⋅ d z d x + y ′ ⋅ 2 y x x ′ ′ = y ′ ′ ⋅ 4 x 2 + y x ′ 2 x ⋅ 2 → y ′ ′ = 1 4 x 2 ( y x x ′ ′ − y x ′ x ) z ⋅ y ′ ′ + y ′ 2 + 3 16 ⋅ y = 0 → x 2 ⋅ 1 4 x 2 ( y x x ′ ′ − y x ′ 2 x ) + 1 2 ⋅ y x ′ x + 3 16 ⋅ y = 0 y x x ′ ′ 4 − 1 4 ⋅ y x ′ x + 1 4 ⋅ y x ′ x + 3 16 ⋅ y = 0 ∣ ⋅ ( 4 ) y x x ′ ′ + 3 4 ⋅ y = 0 The solution will be sought in the form y ( x ) = e k x → y x x ′ ′ = k 2 ⋅ e k x → k 2 ⋅ e k x + 3 16 ⋅ e k x = 0 e k x ⋅ ( k 2 + 3 16 ) = 0 → [ k 1 = − 3 4 ≡ i ⋅ 3 2 k 2 = − − 3 4 ≡ − i ⋅ 3 2 z=x^2\to y'_x=\frac{dy}{dx}=\frac{dy}{dz}\cdot\frac{dz}{dx}=y'\cdot2x\\[0.3cm]
\boxed{y'=\frac{y'_x}{2x}}\\[0.3cm]
y''_{xx}=\frac{d}{dx}\left(y'_x\right)=\frac{d}{dx}\left(y'\cdot 2x\right)=2x\cdot\frac{d\left(y'\right)}{dz}\cdot\frac{dz}{dx}+y'\cdot2\\[0.3cm]
y''_{xx}=y''\cdot 4x^2+\frac{y'_x}{2x}\cdot2\to\boxed{y''=\frac{1}{4x^2}\left(y''_{xx}-\frac{y'_x}{x}\right)}\\[0.3cm]
z\cdot y''+\frac{y'}{2}+\frac{3}{16}\cdot y=0\to\\[0.3cm]
x^2\cdot\frac{1}{4x^2}\left(y''_{xx}-\frac{y'_x}{2x}\right)+\frac{1}{2}\cdot\frac{y'_x}{x}+\frac{3}{16}\cdot y=0\\[0.3cm]
\left.\frac{y''_{xx}}{4}-\frac{1}{4}\cdot\frac{y'_x}{x}+\frac{1}{4}\cdot\frac{y'_x}{x}+\frac{3}{16}\cdot y=0\right|\cdot(4)\\[0.3cm]
\boxed{y''_{xx}+\frac{3}{4}\cdot y=0}\\[0.3cm]
\text{The solution will be sought in the form}\quad y(x)=e^{kx}\to\\[0.3cm]
y''_{xx}=k^2\cdot e^{kx}\to k^2\cdot e^{kx}+\frac{3}{16}\cdot e^{kx}=0\\[0.3cm]
e^{kx}\cdot\left(k^2+\frac{3}{16}\right)=0\to
\left[\begin{array}{l}
k_1=\sqrt{-\displaystyle\frac{3}{4}}\equiv\displaystyle\frac{i\cdot\sqrt{3}}{2}\\[0.4cm]
k_2=-\sqrt{-\displaystyle\frac{3}{4}}\equiv-\displaystyle\frac{i\cdot\sqrt{3}}{2}
\end{array}\right.\\[0.3cm] z = x 2 → y x ′ = d x d y = d z d y ⋅ d x d z = y ′ ⋅ 2 x y ′ = 2 x y x ′ y xx ′′ = d x d ( y x ′ ) = d x d ( y ′ ⋅ 2 x ) = 2 x ⋅ d z d ( y ′ ) ⋅ d x d z + y ′ ⋅ 2 y xx ′′ = y ′′ ⋅ 4 x 2 + 2 x y x ′ ⋅ 2 → y ′′ = 4 x 2 1 ( y xx ′′ − x y x ′ ) z ⋅ y ′′ + 2 y ′ + 16 3 ⋅ y = 0 → x 2 ⋅ 4 x 2 1 ( y xx ′′ − 2 x y x ′ ) + 2 1 ⋅ x y x ′ + 16 3 ⋅ y = 0 4 y xx ′′ − 4 1 ⋅ x y x ′ + 4 1 ⋅ x y x ′ + 16 3 ⋅ y = 0 ∣ ∣ ⋅ ( 4 ) y xx ′′ + 4 3 ⋅ y = 0 The solution will be sought in the form y ( x ) = e k x → y xx ′′ = k 2 ⋅ e k x → k 2 ⋅ e k x + 16 3 ⋅ e k x = 0 e k x ⋅ ( k 2 + 16 3 ) = 0 → ⎣ ⎡ k 1 = − 4 3 ≡ 2 i ⋅ 3 k 2 = − − 4 3 ≡ − 2 i ⋅ 3
Then, the solution has the form
y ( x ) = A 1 ⋅ e k 1 x + A 2 ⋅ e k 2 x ≡ C 1 ⋅ cos ( x 3 2 ) + C 2 ⋅ sin ( x 3 2 ) y(x)=A_1\cdot e^{k_1x}+A_2\cdot e^{k_2x}\equiv C_1\cdot\cos\left(\frac{x\sqrt{3}}{2}\right)+C_2\cdot\sin\left(\frac{x\sqrt{3}}{2}\right)\\[0.3cm] y ( x ) = A 1 ⋅ e k 1 x + A 2 ⋅ e k 2 x ≡ C 1 ⋅ cos ( 2 x 3 ) + C 2 ⋅ sin ( 2 x 3 )
It remains to return to the initial variable and function : z = x 2 → x = z z=x^2\to x=\sqrt{z} z = x 2 → x = z w ( z ) = y ( z ) ⋅ z 1 / 4 w(z)=y(z)\cdot z^{1/4} w ( z ) = y ( z ) ⋅ z 1/4
Conlusion,
w ( z ) = z 1 / 4 ⋅ ( C 1 ⋅ cos ( 3 z 2 ) + C 2 ⋅ sin ( 3 z 2 ) ) \boxed{w(z)=z^{1/4}\cdot\left(C_1\cdot\cos\left(\frac{\sqrt{3z}}{2}\right)+C_2\cdot\sin\left(\frac{\sqrt{3z}}{2}\right)\right)} w ( z ) = z 1/4 ⋅ ( C 1 ⋅ cos ( 2 3 z ) + C 2 ⋅ sin ( 2 3 z ) )
ANSWER
w ( z ) = z 1 / 4 ⋅ ( C 1 ⋅ cos ( 3 z 2 ) + C 2 ⋅ sin ( 3 z 2 ) ) w(z)=z^{1/4}\cdot\left(C_1\cdot\cos\left(\frac{\sqrt{3z}}{2}\right)+C_2\cdot\sin\left(\frac{\sqrt{3z}}{2}\right)\right) w ( z ) = z 1/4 ⋅ ( C 1 ⋅ cos ( 2 3 z ) + C 2 ⋅ sin ( 2 3 z ) )
Comments